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Aug 18, 2022 at 16:45 answer added Ian Agol timeline score: 4
Jul 20, 2022 at 9:14 vote accept dennis
Jul 19, 2022 at 15:23 comment added Ryan Budney You want to smooth $f^2$, that way you do not need to modify the function near $\Sigma$.
Jul 19, 2022 at 10:44 comment added dennis @RyanBudney If the smoothed function $\tilde{f}$ is used, then $f(x)$ and $\tilde f(x)$ are not necessarily equal, so I don't think $\Sigma=\tilde f^{-1}(0)$.
Jul 16, 2022 at 15:40 comment added Moishe Kohan For incomplete Riemannian manifolds the answer is negative.
Jul 16, 2022 at 8:36 comment added Ryan Budney The function $f^2$ is smooth near the manifold $\Sigma$, but it could potentially have some non-smooth points far from $\Sigma$. These could be smoothed-away using bump functions.
Jul 16, 2022 at 0:55 comment added Ryan Budney Thanks, ThiKu has answered your question provided the Riemann manifold is complete.
Jul 15, 2022 at 23:50 history edited dennis CC BY-SA 4.0
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Jul 15, 2022 at 21:58 vote accept dennis
Jul 15, 2022 at 23:49
Jul 15, 2022 at 4:45 comment added Moishe Kohan In your setting, one can make the submanifold to consist of regular points of the defining function (since orientable 3-manifolds have trivial tangent bundle).
Jul 14, 2022 at 23:10 comment added Ryan Budney Using the word "implicitly" are you referring to the implicit function theorem, i.e. do you want $0$ to be a regular value of $f$?
Jul 14, 2022 at 23:04 answer added ThiKu timeline score: 10
Jul 14, 2022 at 22:51 history edited dennis CC BY-SA 4.0
added 12 characters in body; edited title
Jul 14, 2022 at 20:54 history edited dennis
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S Jul 14, 2022 at 20:48 review First questions
Jul 14, 2022 at 21:18
S Jul 14, 2022 at 20:48 history asked dennis CC BY-SA 4.0